Question

The equation of a particle executing simple harmonic motion is given by \[ x = \sin \pi \left( t + \frac{1}{3} \right) \, {m}. \] At \( t = 1 \, {s} \), the speed of particle will be (Given \( \pi = 3.14 \)):

• A. \( 0 \, {cm s}^{-1} \)
• B. \( 157 \, {cm s}^{-1} \)
• C. \( 272 \, {cm s}^{-1} \)
• D. \( 314 \, {cm s}^{-1} \)

Hint:

For simple harmonic motion, the velocity is the time derivative of the displacement equation. Use the relation \( v = A\omega \cos(\omega t + \phi) \) to find the speed of the particle.

Answer 1

The Correct Option is B

Step 1: The equation for displacement of a particle in simple harmonic motion is given by \[ x = A \sin(\omega t + \phi), \] where \( A \) is the amplitude, \( \omega \) is the angular frequency, and \( \phi \) is the phase constant. 
Step 2: Given, \( x = \sin \pi \left( t + \frac{1}{3} \right) \), so \( A = 1 \, {m} \) and \( \omega = \pi \). 
Step 3: The velocity \( v \) is the time derivative of displacement: \[ v = \frac{dx}{dt} = A \omega \cos(\omega t + \phi). \] 
Step 4: At \( t = 1 \, {s} \), \[ v = \pi \cos\left( \pi \times 1 + \pi \times \frac{1}{3} \right) = \pi \cos\left( \frac{4\pi}{3} \right) = \pi \times (-\frac{1}{2}) = -\frac{\pi}{2} \, {m/s}. \] The speed \( |v| \) is: \[ |v| = \frac{\pi}{2} \, {m/s} = 157 \, {cm/s}. \]